Question

Multiply $\frac{-8}{21}{x}^2{y}^3$ by$\frac{-7}{16}x{y}^2$ and verify your result for $x=3$ and $y=2$.

Answer 1

We have to multiply, \(\dfrac{-8}{21}x^{2}y^{3}\) by \(\dfrac{-7}{16}xy^{2}\)

$=\frac{-8}{21}{x}^2{y}^3\times \frac{-7}{16}x{y}^2$

$=\frac{-8}{21}\times \frac{-7}{16}{x}^{2}x{y}^3{y}^2$

$=\frac{1}{3}\times \frac{1}{2}{x}^{2+1}{y}^{3+2}$ [Since, $a^m\times a^n=a^{m+n}$]

$=\frac{1}{6}{x}^3{x}^5$

$\therefore \frac{-8}{21}{x}^2{y}^3\times \frac{-7}{16}x{y}^2=\frac{1}{6}{x}^3{x}^5$

Now, lets verify the result for $x=3$ and $y=2$

LHS $=\frac{-8}{21}{x}^2{y}^3$ by$\frac{-7}{16}x{y}^2$

$=\frac{-8}{21}\left(3{)}^2\right(2{)}^3\times \frac{-7}{16}\left(3\right)(2{)}^2$

$=\frac{-8}{21}\times \frac{-7}{16}\times \left(3{)}^2\right(2{)}^3\left(3\right)(2{)}^2$

$=\frac{1}{3}\times \frac{1}{2}\times 9\times 8\times 3\times 4$

$=3\times 4\times 3\times 4$

$=144$

RHS $=\frac{1}{6}{x}^3{x}^5$

$=\frac{1}{6}\times 3^3{2}^5$

$=\frac{27\times 32}{6}$

$=9\times 16$

$=144$

Therefore, LHS $=$ RHS.

Hence, Verified.